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Not sure the implication holds: if $X_0=0$ and $X_t=1$ for every positive $t$ then both sums are $1$ for every partition $\pi$ hence the first condition (boundedness) is met while the second (convergence to zero) is not.

Edit: To get a less degenerate Gaussian process $(X_t)_{t\geqslant0}$, try $X_0=0$ and $X_t=\xi$ for every positive $t$, for some nonzero normal random variable $\xi$ (note that our previous example was a Gaussian process as well).

I do not understand how a deterministic constant function can follow a normal distribution. Also I'm not sure, whether $(\xi, \xi)$ follows a 2-dimensional gaussian Distribution, as it has to, to be a gaussian process.
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Maximilian KönigDec 16 '12 at 19:51

All this points at the fact that you could check the definition of being a normal vector. You shall see that Dirac distributions are normal.
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DidDec 16 '12 at 19:58

Thank you very much. Especially that last thing helped me much for my understanding.
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Maximilian KönigDec 16 '12 at 20:36